Assume that \(S\cong A_{\ell }\) , \(\ell \geqslant 5\) . Then \(Y\)  is one of the groups
listed in Example 2.6.(a)–(c) of [1]}. If \(Y\)  is one of the groups
in Example 2.6.(b)–(c), then \(\ell \leqslant 10\)  and \(\Phi _{ef}^{\ast }(p)^{1/2}\)  is one of the primes 5 or 7, as \(n\geqslant 3\)  (see Tables 2–3
in [1]}). Then \(\Phi _{ef}^{\ast }(p)\)  is either \(5^{2}\)  or \(7^{2}\) .
Actually, only the former is admissible and occurs for \((ef,p)=(4,7)\)  by
[3]}, Theorem 3. Then \(Z_{2}.A_{7}\trianglelefteq Y\)  by Tables 2–3
in [1]}, since \(p\mid \left|Y\right|\) , and hence
\(Z_{2}.A_{7} \unlhd Y \leqslant Z_{2}.S_{7} \)  by [5]}.
Suppose that \(Y\)  is one of the groups in Example 2.6.(a). Then \(V_{n}(q)\) , \(q=p\) , is the fully deleted permutation module for \(S\)  and \(A_{\ell }\leqslant Y\leqslant S_{\ell }\times Z_{p-1}\) , where \(\ell =n+\varepsilon \) , \(\varepsilon =1\)  or 2 according as \(p\)  does not divide or does divide \(n\) 
respectively. Let \(u\)  be any prime divisor of \(\Phi _{ef}^{\ast }(p)\) . Then \(u^{i}\mid \left|Y\right|\) , and hence \(u^{i}\mid \left|S\right|\) , for some \(i\geqslant 1\) . Then \(i< \frac{\ell -1}{u}\)  by
[6]}, Exercise 2.6.8. On the other hand, \(u=1+te\) , for some \(t\geqslant 1\) , by [7]}, Proposition 5.2.15.(i). Thus \(1+i(1+te)<n+\varepsilon \) , where \(e=n,n-1\)  or \(n-2\) . Then \(e \ne n\) , \(i=1\) , and either \(t=1\) , or \(\varepsilon =2\) , \(e=1\) , \(n=p=3\)  and \(t=2\) . The latter implies \(u=p=3\) , and hence a contradiction, whereas the former yields \((e,q,u)=(4,7,5)\)  by [3]}, Theorem 3, since \(\Phi _{e}^{\ast }(q)=(e+1)^{2}\) . Then \((\ell ,n,\varepsilon ) =(7,6,1)\) , since \(\ell =n+\varepsilon \) , \(e=n-1,n-2\)  and \(7 \mid \left|Y\right|\) . However, \(A_{7}\)  does not afford an irreducible representation on \(V_{4}(7)\)  by [5]}, and this case is excluded.
Assume that \(S\)  is sporadic. Then \(S\)  is one of the groups listed in Table 5
of [1]}. Note that, the order of each of these groups is divisible by
exactly one primitive prime divisor of \(p^{e}-1\) , with \(e=n,n-1\)  or \(n-2\) . Thus \(\Phi _{ef}^{\ast }(p)\)  is the square of a prime, and hence \((ef,p,u)=(4,7,5)\)  or \((5,3,11)\)  by [3]},
Theorem 3. Only the latter yields a group that is listed in Table 5 and this is \(Y^{\prime }\cong M_{11}\) . Then \(Y\cong M_{11}\) , since \(\mathrm {Out(}M_{11}\mathrm {)}=1\)  by [5]}.
Assume that \(S\)  is a Lie type simple group in characteristic different from \(p\) . Then \(Y\)  can be determined from Tables 7–8 of [1]}. As in the
sporadic case, the order of the groups in these two tables is divisible by
exactly one primitive prime divisor of \(p^{e}-1\) , with \(e=n,n-1\)  or \(n-2\) . Then \(\Phi _{ef}^{\ast }(p)\)  is the square of a prime and hence \((ef,p,u)=(4,7,5)\)  or \((5,3,11)\)  as above. If \(S\)  is listed in Table 7, then
\(S\cong Sp_{4}(3)\) , but no groups \(Y\)  occur with order divisible by 7. Hence, assume \(S\)  is one of the groups listed in Table 8. If \(u=5\) , then \(e=4\)  and it is easy to see that \(S\)  is
isomorphic to one of the groups \(PSp_{4}(3)\) , or \(PSL_{2}(9)\) . However, there are no corresponding \(Y\)  with order divisible by 7. Thus, \(u=11\) , \(u=2e+1\) 
and hence \(s=11\)  and \(S\cong PSL_{2}(11)\) . Then \(Y\cong PSL_{2}(11)\) 
by [5]}.
Assume that \(S\)  is a Lie type simple group in characteristic \(p\) . Then \(Y\)  is
one of the groups in Example 2.8, and hence in Table 6, of [1]}. Let
\(GF(q_{0})\)  be a subfield of \(GF(q)\)  for which \(Y\leqslant GL_{n}(q_{0})\circ Z_{q-1}\) , and \(Y\cap GL_{n}(q_{0})\)  cannot be realized modulo scalar over a
proper subfield of \(GF(q_{0})\) . Let \(u\)  be a prime divisor of \(\Phi _{ef}^{\ast }(p)\) , then \(u\)  is a primitive prime divisor of \(p^{ef}-1\)  and
this forces \(q_{0}=q\)  by [7]}, Proposition 5.2.15.(ii).
Assume that \(n=8\) , \(e=6\)  and \(Y^{(\infty )}\cong SL_{2}(q^{3})\) . Then \(\left[ SL_{8}(q):Y\right] \)  divides
\(\left[ SL_{8}(q):Y^{(\infty )}\right] =q^{25}\prod _{i=2,i\ne 6}^{8}(q^{i}-1)\) 
and hence \(\Phi _{6f}^{\ast }(p)^{1/2}\)  does not divide \(\left[ SL_{8}(q):Y\right] \) , a contradiction. A similar reasoning rules out the remaining
cases in Table 6.
